Concept

Traceback — where it appears

Walking a filled table backwards from its answer to recover the choices that produced it, which reads cells a rolling frontier has already released. It needs cells a rolling frontier has already released, which is the reason a linear-space method computes the table twice rather than once.

Named by 7 essays across 3 fields — each of them below, with the objects they name alongside it.

0%25%50%75%100%11.522.533.545what a matching character is worthreported as the shared region, of the longer sequencea random pair scores zero hereunit cost, alphabet acgtthe region found: 13 characters at the left, 34 at the right

The zero that moves the answer out of the corner

One extra term in the recurrence — a floor at zero — and the answer stops being in the last cell. It becomes a maximum over all 1,040 of them, the traceback's starting point is a search, and the whole mode is meaningless unless a randomly matched pair of characters scores negative on average. That last condition is on the scoring scheme, not on the sequences.

tables · Cost
agcacacggatcagccagggagta09101112131415161718192021229110101212141416171819192122101021011121314151718191920221110113101212141516171920192112121012411131314151718202119131312111351214131416181921211414131312146131413151719202115141514131314714151316181921161614161514141571416141719191717171517161515168151615182018171818151816161617816171519one unit = one subproblem given a value495 cells across three tables, 980 transitions

A cell that has to know where it is

A gap of four characters is usually one event, not four. No recurrence over a single table can charge it that way, because the price of a gap character depends on how the cell above it was reached and a cell holding one number has thrown that away. The repair is three tables, and it costs exactly three times the cells.

tables · Cost
executionintention87777776569888888765one unit = one subproblem given a value100 cells computed, 20 held at once

The table nobody has to keep

A million-cell table, computed cell for cell in the same order, holding two thousand cells at its peak instead of a million. The saving is exactly (n+1)/2, it costs nothing on any operation counter, and what it buys is paid for with the one thing the table was for.

space · Table
gactacgatgattacagt0123456789101234567821012345673211123456432212345554332123456543321234765443222387655432339876554333one unit = one subproblem given a value100 cells, 100 held at once

A distance that is a path through a grid

How far apart two strings are is a shortest-path problem on a grid whose every edge is drawn by the recurrence — and finding a string in a text costs 4,988 character comparisons where measuring how far it is from one costs 96,000.

text · Distance
executionintention0123456789112345667822234567773333455678434345667854444567776555555678766666656787777776569888888765one unit = one subproblem given a value110 cells for one divide step, 30 held

The alignment that fits in one line

Compute the table twice and hold three rows of it. The factor of two is a geometric series and is predicted exactly; measured, it comes down from 2.269 to 2.052 as the strings grow, and the peak is 3(m+1) cells on the nose.

space · Distance
123456123456cost to open a gapcost to extend a gapintentionexecution571 settingsinte---ntion---execution3 settingsinte-ntion-execution1 setting-intentionexec-ution1 settingintention against execution, affine costseach colour is one optimal alignment

The parameter plane has few answers

Sweep the cost of opening a gap against the cost of extending one over five hundred and seventy-six settings, and the optimal alignment of intention against execution takes four values — one of them at 571 of the settings. Under a linear model the plane divides into three wedges through the origin, because doubling every cost changes nothing and only the ratio is a parameter. Tuning an aligner is choosing a region, and most of the plane is one.

tables · Cost
rounded to whole bitsunroundeda resample, 400 near pairs0.860.9340 pairs at stay 0.90.820.928 pairs at stay 0.90.710.88400 pairs at stay 0.70.680.86400 pairs at stay 0.50.620.790.5: no prediction200 test pairs, 16 directionsdashed: a coin flip

The ties a rounded matrix makes

Measure how far each optimal alignment is from a tie — the smallest change to any one cost that makes another alignment win — and it predicts which alignments a refitted substitution matrix will move. A resample of the same corpus moves 30 of the 63 test alignments that sit on a tie and 3 of the other 137. A matrix fitted to a different divergence moves alignments far from a tie as well, and the prediction weakens to a chance of 0.62. And a third of the alignments were on a tie only because the matrix was rounded to whole bits — fitted without rounding, 15 of 200 are, and every prediction improves.

tables · Cost

Named alongside it

The objects these essays reach for when they reach for this one.

AlignmentCost modelDynamic programmingEdit distanceSubproblemHonest limitParameter choiceTrade offAffine gapAuxiliary spaceOptimalityPeak space

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